\left. \begin{array} { l } { f ( x ) = \operatorname { log } _ { [ x +

1.	\left. \begin{array} { l } { f ( x ) = \operatorname { log } _ { [ x + \frac { 1 } { 2 } ] } \| x ^ { 2 } - 5 x + 6 \| \text { is } } \\ { [ \frac { 3 } { 2 } , 2 ) \cup ( 2,3 ) \cup ( 3 , \infty ) } \\ { ( b ) [ \frac { 3 } { 2 } , \infty ) } \\ { ( \frac { 1 } { 2 } , \infty ) } \\ { ( c ) [ \frac { 1 } { 2 } , \infty ) } \end{array} \right.
Answer» For log function , the domain should be such that the base should be 0 , or 1 and the number should > 0 [x+1/2] .. is the greatest integer function.. and for this function not to equal will 0 or 1 The minimum value of [x+1/2] = 2 => X +1/2 ≥2 => x ≥ 2-1/2 ≥ 3/2.... also.=> \|x²-5x+6\| ≠ 0=> \|(x-2)(x-3)\| ≠ 0 so, 2 and 3 should also be excluded from the domain. The final domain will be [3/2) U (2,3) U (3,∞) option A

\left. \begin{array} { l } { f ( x ) = \operatorname { log } _ { [ x + \frac { 1 } { 2 } ] } | x ^ { 2 } - 5 x + 6 | \text { is } } \\ { [ \frac { 3 } { 2 } , 2 ) \cup ( 2,3 ) \cup ( 3 , \infty ) } \\ { ( b ) [ \frac { 3 } { 2 } , \infty ) } \\ { ( \frac { 1 } { 2 } , \infty ) } \\ { ( c ) [ \frac { 1 } { 2 } , \infty ) } \end{array} \right.

Answer»

For log function , the domain should be such that the base should be 0 , or 1

and the number should > 0

[x+1/2] .. is the greatest integer function.. and for this function not to equal will 0 or 1 The minimum value of [x+1/2] = 2

=> X +1/2 ≥2 => x ≥ 2-1/2 ≥ 3/2....

also.=> |x²-5x+6| ≠ 0=> |(x-2)(x-3)| ≠ 0

so, 2 and 3 should also be excluded from the domain.

The final domain will be

[3/2) U (2,3) U (3,∞)

option A